Python|按子列表的大小对列表列表进行排序

给定一个列表列表，任务是根据子列表的大小对列表进行排序。让我们讨论一些方法来做同样的事情。方法#1：使用排序

Python3

# Python code to demonstrate
# sort list of list
# on the basis of size of sublist
 
ini_list = [[1, 2, 3], [1, 2], [1, 2, 3, 4],
                [1, 2, 3, 4, 5], [2, 4, 6]]
 
# printing initial ini_list
print ("initial list", str(ini_list))
 
# sorting on basis of size of list
ini_list.sort(key = len)
 
# printing final result
print("final list", str(ini_list))

输出：初始列表 [[1, 2, 3], [1, 2], [1, 2, 3, 4], [1, 2, 3, 4, 5], [2, 4, 6]] 最终列表 [[1, 2], [1, 2, 3], [2, 4, 6], [1, 2, 3, 4], [1, 2, 3, 4, 5]]

方法 #2：使用 lambda

Python3

# Python code to demonstrate
# sort list of list
# on the basis of size of sublist
 
ini_list = [[1, 2, 3], [1, 2], [1, 2, 3, 4],
                [1, 2, 3, 4, 5], [2, 4, 6]]
 
# printing initial ini_list
print ("initial list", str(ini_list))
 
# sorting on basis of size of list
ini_list.sort(key = lambda x:len(x))
 
# printing final result
print("final list", str(ini_list))

输出：初始列表 [[1, 2, 3], [1, 2], [1, 2, 3, 4], [1, 2, 3, 4, 5], [2, 4, 6]] 最终列表 [[1, 2], [1, 2, 3], [2, 4, 6], [1, 2, 3, 4], [1, 2, 3, 4, 5]]

方法#3：使用排序

Python3

# Python code to demonstrate
# sort list of list
# on the basis of size of sublist
 
ini_list = [[1, 2, 3], [1, 2], [1, 2, 3, 4],
                [1, 2, 3, 4, 5], [2, 4, 6]]
 
# printing initial ini_list
print ("initial list", str(ini_list))
 
# sorting on basis of size of list
result = sorted(ini_list, key = len)
 
# printing final result
print("final list", str(result))

输出：初始列表 [[1, 2, 3], [1, 2], [1, 2, 3, 4], [1, 2, 3, 4, 5], [2, 4, 6]] 最终列表 [[1, 2], [1, 2, 3], [2, 4, 6], [1, 2, 3, 4], [1, 2, 3, 4, 5]]